JIPMAT Polygons PYQs

2 solved Polygons previous year questions (PYQs) from JIPMAT past year papers — attempt each and check the answer.

Q1:jipmat 2025QA › PolygonsMediumQA · MCQ

The cost of fencing of an equilateral triangular park and a square park is the same. If the area of the triangular park is

16\sqrt{3}\text{ m}^2

then the length of the diagonal of the square park is:

A $6\sqrt{3}\text{ m}$
B $8\sqrt{3}\text{ m}$
C $8\sqrt{2}\text{ m}$
D $6\sqrt{2}\text{ m}$

Pick an option to attempt

The Setup: The fencing costs are identical, meaning their perimeters perfectly match each other's energy (they are exactly equal). We need to find the side of the triangle, use it to get the perimeter, transfer that perimeter to the square, and finally find the square's diagonal. Step 1: Find the side length of the equilateral triangle (

a

). Area formula:

\frac{\sqrt{3}}{4}a^2=16\sqrt{3}

. Cancel out

\sqrt{3}

\frac{a^2}{4}=16

a^2=64 \implies a=8\text{ m}

. Step 2: Calculate the perimeter of the triangle.

P=3 \times 8=24\text{ m}

Step 3: Find the side of the square (

s

). Since perimeters are equal:

4s=24 \implies s=6\text{ m}

Step 4: Find the diagonal of the square using the formula

d=s\sqrt{2}

d=6\sqrt{2}\text{ m}

Final Answer:

6\sqrt{2}

Q2:jipmat 2025QA › PolygonsEasyQA · MCQ

Arrange the following in descending order based on their perimeters: A. A square with an area of 36 sq. cm. B. An equilateral triangle with a side of 9 cm. C. A rectangle with 10 cm as length and 40 sq. cm as area. D. A circle with radius of 4 cm.

AC > A > B > D
BD > A > B > C
CC > B > A > D
DC > B > D > A

Pick an option to attempt

The Setup: We need to calculate the perimeter (or circumference) for each shape to see who is really built different, then rank them from largest to smallest. Step 1: Calculate A (Square).

\text{Area}=s^2=36 \implies s=6\text{ cm}

\text{Perimeter}=4s=4 \times 6=24\text{ cm}

. Step 2: Calculate B (Equilateral Triangle). Side is

9\text{ cm}

\text{Perimeter}=3s=3 \times 9=27\text{ cm}

. Step 3: Calculate C (Rectangle).

\text{Area}=L \times W \implies 40=10 \times W \implies W=4\text{ cm}

\text{Perimeter}=2(L+W)=2(10+4)=2(14)=28\text{ cm}

. Step 4: Calculate D (Circle). Radius

r=4\text{ cm}

\text{Circumference}=2\pi r=2 \times 3.1415 \times 4 \approx 25.13\text{ cm}

. Step 5: Rank them in descending order (biggest first).

C(28) > B(27) > D(25.13) > A(24)

. Final Answer: C > B > D > A

JIPMAT Polygons — PYPs

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